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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Mon, 2 Dec 2024 16:48:34 -0800
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On 12/2/2024 4:43 PM, Ross Finlayson wrote:
> On 12/02/2024 04:32 PM, Jim Burns wrote:
>> On 12/2/2024 9:28 AM, WM wrote:
>>> On 02.12.2024 12:53, FromTheRafters wrote:
>>
>>>> [...]
>>>
>>> Infinite endsegments contain an infinite set each,
>>> infinitely many elements of which
>>>  are in the intersection.
>>
>> Yes to:
>> ⎛ regarding finite.cardinals,
>> ⎜ for  each  end.segment   E(k)
>> ⎜ there is a subset S such that
>> ⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
>>
>> No to:
>> ⛔⎛ regarding finite.cardinals,
>> ⛔⎜ ⮣ there is a subset S such that ⮧
>> ⛔⎜ ⮤ for  each  end.segment   E(k) ⮠
>> ⛔⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
>>
>> A quantifier shift tells you (WM) what you (WM) _expect_
>>   but a quantifier shift is untrustworthy.
>>
>>> An empty intersection cannot come before
>>> an empty endsegment has been produced by
>>> losing one element at every step.
>>
>> No.
>> Because see below.
>>
>>> E(1), E(2), E(3), ...
>>> and
>>> E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
>>> are identical for every n and in the limit
>>> because
>>> E(1)∩E(2)∩...∩E(n) = E(n).
>>
>> They are
>> identical COUNTER.EXAMPLES to what you expect.
>>
>> ----
>>> An empty intersection cannot come before
>>> an empty endsegment has been produced by
>>> losing one element at every step.
>>
>> No.
>> For the set of finite cardinals,
>> EVEN IF NO END.SEGMENT IS EMPTY,
>>   the intersection of all end segments is empty.
>>
>> ⎛ The set of finite.cardinals holds
>> ⎜  only finite.cardinals.
>> ⎜
>> ⎜ Each finite.cardinal is finite.
>> ⎜
>> ⎜ For each finite.cardinal,
>> ⎜  only finitely.many finite.cardinals are ≤ it.
>> ⎜
>> ⎜ For each finite.cardinal,
>> ⎜  only end.segments which start ≤ it
>> ⎜ hold it.
>> ⎜
>> ⎜ For each finite.cardinal,
>> ⎝  only finitely.many end.segments hold it.
>>
>> ⎛ The set of finite cardinals holds
>> ⎜  all finite.cardinals.
>> ⎜
>> ⎜ Each finite.cardinal is followed by
>> ⎜  another finite.cardinal.
>> ⎜
>> ⎜ No finite.cardinal is last.
>> ⎜
>> ⎜ The set of finite cardinals has
>> ⎜  a subset (itself) which is not two.ended.
>> ⎜
>> ⎜ The set of finite cardinals is infinite.
>> ⎜
>> ⎜ Each finite.cardinal starts an end.segment.
>> ⎜
>> ⎝ There are infinitely.many end.segments.
>>
>> ⎛ For each finite.cardinal,
>> ⎜  only finitely.many end.segments hold it.
>> ⎜
>> ⎜ There are infinitely.many end.segments.
>> ⎜
>> ⎜ For each finite.cardinal,
>> ⎜  not all end.segments hold it.
>> ⎜
>> ⎜ For each finite.cardinal,
>> ⎜  the intersection doesn't hold it
>> ⎜
>> ⎜ EVEN IF NO END.SEGMENT IS EMPTY,
>> ⎜⎛ For each finite.cardinal,
>> ⎜⎜  only finitely.many end.segments hold it.
>> ⎜⎜ For each finite.cardinal,
>> ⎜⎜  only finitely.many end.segments hold it.
>> ⎜⎜ For each finite.cardinal,
>> ⎜⎝  the intersection doesn't hold it
>> ⎜
>> ⎜ EVEN IF NO END.SEGMENT IS EMPTY,
>> ⎝  the intersection of all end segments is empty.
>>
>>
> 
> The usual idea of wrestling with a pig is
> that you both get dirty, and the pig likes it.
> 
> 
> Quit letting that pig dirty things.

Humm... Indeed. I have heard this before:

Never get into an argument with a pig. It will drag you down to its 
piggish low level and beat you with experience... ;^o